= (c + a,d + b) = (c, d) + (a,b). (0, 0) + (a,b) = (0 + a,0 + b) = (a,b) (a,b) + ( a, b) = (a + a,b + b) = (0,0) = 0.

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Sydney Parker
6 years ago
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1 Question Show that R 2 is a vector space Solution We need to check each and every axiom of a vector space to know that it is in fact a vector space A1: Let (a,b), (c,d) R 2 Then (a,b) + (c,d) (a + c,b + d) R 2 Therefore A1 holds A2: Let (a,b), (c,d) R 2 Then (a,b) + (c,d) (a + c,b + d) (c + a,d + b) (c, d) + (a,b) Therefore A2 holds A3: Let (a,b), (c,d), (e, f) R 2 Then (a, b) + ((c,d) + (e,f)) (a,b) + (c + e, d + f) (a + (c + e),b + (d + f)) ((a + c) + e, (b + d) + f) (a + c,b + d) + (e,f) ((a,b) + (c, d)) + (e,f) Therefore A3 holds A4: Our claim is that the vector (0, 0) works Let (a,b) R 2 Then (0, 0) + (a,b) (0 + a,0 + b) (a,b) Therefore, A4 holds A5: Let (a,b) R 2 Then we need to find (a,b) Our claim is that ( a, b) works (a,b) + ( a, b) (a + a,b + b) (0,0) 0 M1: Let k R, (a,b) R 2 Then k(a,b) (ka,kb) R 2 Therefore M1 holds M2: Let k R, (a,b), (c,d) R 2 Then, k((a,b) + (c,d)) k(a + c,b + d) (k(a + c),k(b + d)) (ka + kc, kb + kd) (ka,kb) + (kc,kd) k(a, b) + k(c,d)

2 Therefore M2 holds M3: Let k,m R, (a,b) R 2 Then (k + m)(a, b) ((k + m)a,(k + m)b) (ka + ma,kb + mb) (ka,kb) + (ma,mb) k(a,b) + m(a,b) Therefore M3 holds M4: Let k,m R, (a,b) R 2 Then k(m(a,b)) k(ma,mb) (k(ma),k(mb)) ((km)a,(km)b) (km)(a, b) Therefore M4 holds M5: Let (a,b) R 2 Then, 1(a,b) (1a,1b) (a,b) Therefore M5 holds

3 Question Show that M, the set of all 2 2 matrices, is a vector space Solution We need to check each and every axiom of a vector space to know that it is in fact a vector space A1: Let 1,2 b1,1 b, 1,2 b 2,1 b Then which is also a 2 2 matrix Therefore 1,2 A2: Let 1,2 b1,1 b a1,1 + b 1,1 a 1,2 + b 1,2, b 2,1 b a 2,1 + b 2,1 a + b b1,1 b b 2,1 b 1,2 b1,1 b, 1,2 c1,1 c, 1,2 b 2,1 b c 2,1 c Then ( ) 1,2 b1,1 b c1,1 c 1,2 b 2,1 b c 2,1 c () b1,1 + c + 1,1 b 1,2 + c 1,2 b 2,1 + c 2,1 b + c a1,1 + b 1,1 + c 1,1 a 1,2 + b 1,2 + c 1,2 a 2,1 + b 2,1 + c 2,1 a + b + c () a1,1 + b 1,1 a 1,2 + b 1,2 c1,1 c a 2,1 + b 2,1 a + b c 2,1 c ( ) 1,2 b1,1 b c1,1 c b 2,1 b c 2,1 c A3: Let 1,2 b1,1 b, 1,2 b 2,1 b 1,2 Then b1,1 b b 2,1 b a1,1 + b 1,1 a 1,2 + b 1,2 a 2,1 + b 2,1 a + b b1,1 + a 1,1 b 1,2 + a 1,2 b 2,1 + a 2,1 b + a b1,1 b 1,2 + b 2,1 b 1,2 A4: The vector 0 is the zero matrix A5: Let A, since for any 1,2, 1,2 1,2 a1,1 a, and define A to be the matrix 1,2 Then a 2,1 a 1,2 which is the zero vector 0 as required a1,1 a + 1,2 a1,1 a 1,1 a 1,2 a 1,2 a 2,1 a a 2,1 a,

4 M1: Let r R, and let 1,2 Let r 1,2 ra1,1 ra 1,2 ra 2,1 ra M2: Let r R, and let 1,2 b1,1 b, 1,2 b 2,1 b Then ( ) r 1,2 b1,1 b b 2,1 b () a1,1 + b r 1,1 a 1,2 + b 1,2 a 2,1 + b 2,1 a + b r(a1,1 + b 1,1 ) r(a 1,2 + b 1,2 ) r(a 2,1 + b 2,1 ) r(a + b ) ra1,1 + rb 1,1 ra 1,2 + rb 1,2 ra 2,1 + rb 2,1 ra + rb ra1,1 ra 1,2 rb1,1 rb ra 2,1 ra rb 2,1 rb M3: Let r,s R, and let 1,2 Then (r + s) 1,2 (r + s)a1,1 (r + s)a 1,2 (r + s)a 2,1 (r + s)a ra1,1 + sa 1,1 ra 1,2 + sa 1,2 ra 2,1 + sa 2,1 ra + sa ra1,1 ra 1,2 sa1,1 sa ra 2,1 ra sa 2,1 sa M4: Let r,s R, and let 1,2 Then ( r s 1,2 ) () sa1,1 sa r 1,2 sa 2,1 sa r(sa1,1 ) r(sa 1,2 ) r(sa 2,1 ) r(sa ) (rs)a1,1 (rs)a 1,2 (rs)a 2,1 (rs)a (rs) 1,2 M5: For any 1,2, 1 1,2 1a1,1 1a 1,2 1,2 1a 2,1 1a

5 Question Determine if the set V of solutions of the equation 2x 3y + z 1 is a vector space or not Determine which axioms of a vector space hold, and which ones fail The set V (together with the standard addition and scalar multiplication) is not a vector space In fact, many of the rules that a vector space must satisfy do not hold in this set What follows are all the rules, and either proofs that they do hold, or counter examples showing they do not hold A1: u,v V u + v V (closure under addition) Let u,v V That is, u and v are solutions to the equation 2x 3y + z 1 More specifically, u and v are triples in R 3, say u (u 1, u 2,u 3 ) and v (v 1,v 2,v 3 ), such that 2u 1 3u 2 + u 3 1 and 2v 1 3v 2 + v 3 1 A1 does not hold here For instance, take u (0, 0, 1) and v (1, 0, 1) (both are in V since both are solutions to 2x 3y +z 1) Then u+v (1, 0,0), but 2(1) 3(0)+0 2 1, and therefore u+v is not a solution to 2x 3y +z 1, showing that u + v V A2: Associativity holds since the real numbers are associative Let u (u 1, u 2,u 3 ),v (v 1,v 2,v 3 ),w (w 1,w 2,w 3 ) V Then u + (v + w) (u 1,u 2,u 3 ) + ((v 1,v 2,v 3 ) + (w 1, w 2,w 3 )) (u 1,u 2,u 3 ) + (v 1 + w 1,v 2 + w 2,v 3 + w 3 ) (u 1 + v 1 + w 1,u 2 + v 2 + w 2,u 3 + v 3 + w 3 ) (u 1 + v 1,u 2 + v 2, u 3 + v 3 ) + (w 1,w 2,w 3 ) ((u 1,u 2,u 3 ) + (v 1,v 2,v 3 )) + (w 1,w 2,w 3 ) (u + v) + w A3: Commutativity holds (similar to associativity above) just since the real numbers are commutative Let u (u 1,u 2,u 3 ),v (v 1,v 2,v 3 ) V Then u + v (u 1,u 2,u 3 ) + (v 1,v 2,v 3 ) (u 1 + v 1,u 2 + v 2,u 3 + v 3 ) (v 1 + u 1,v 2 + u 2,v 3 + u 3 ) (v 1,v 2,v 3 ) + (u 1,u 2,u 3 ) v + u A4: The requirement of a zero vector fails since there is only one possibility for a zero vector using the standard addition: the vector (0, 0, 0), which is not a solution to the equation 2x 3y + z 1 (since 2(0) 3(0) + (0) 0) Therefore, A4 fails A5: The requirement of additive inverses fails as well For instance, (0, 0,1) is an element of V (as mentioned in A1 above) The additive inverse of (0,0, 1), using the standard addition, would be (0, 0, 1) However, (0, 0, 1) is not a solution to the equation 2x 3y + z 1 Therefore (0, 0, 1) V In fact, the following proposition shows that for every v V, v V : Proposition If v V, then v V Proof Let v (v 1,v 2,v 3 ) V Then v is a solution to the equation 2x 3y + z 1, that is, 2v 1 3v 2 + v 3 1 Plugging in v ( v 1, v 2, v 3 ) into the equation, we get: 2( v 1 ) 3( v 2 ) + ( v 3 ) 2v 1 + 3(v 2 ) v 3 ( 1)(2v 1 3v 2 + v 3 ) ( 1)(1) (since 2v 1 3v 2 + v 3 1) 1 1

6 Therefore, v ( v 1, v 2, v 3 ) is not a solution to 2x 3y + z 1, and is therefore not in V One might say the above proposition proves this space TOTALLY fails A5, since for EVERY v V, the additive inverse of v, that is v, is not in V Note the above also produces an infinite number of counter examples to M 1 (below) For every v V, ( 1)v V M1: Closure under multiplication does not hold For example, take 2 R, and (0, 0,1) V Then 2(0, 0,1) (0,0, 2), but (0, 0, 2) is not a solution to the equation 2x 3y + z 1 (since 2(0) 3(0) + (2) 2 1) Therefore, the space is not closed under scalar multiplication M2, M3, M4, and M5 all hold, for the same reasons they hold in R 3 with the standard addition, proofs similar to the above for A2 and A3

7 Question Let V be the set V {red, blue}, that is, V is the finite set consisting of just the two elements red and blue Define an addition on this set as follows: Red Blue Red Red Blue Blue Blue Red For any real number r, for any v V, define the scalar multiplication r v as r v { red if r 0, blue otherwise Show which axioms of a vector space hold, and which fail for this set V together with these operations and Solution A1: Let u,v V Then by the definition of the addition, u + v is either red or blue, and is thus again in V Therefore V is closed under addition (A1 holds) A2: Let u,v,w V To show this holds, we could check every possibility for u, v, and w one by one, but this would take a while To do it faster, I ll be a little smarter about what I check If u red, then u (v w) red (v w) v w (red x x) (red v) w (red v v) (u v) w, which is what was needed to be shown If v red, then similarly, u (v w) u (red w) u w (u red) w (u red u) (u v) w, which is what was needed to be shown If w red, then, u (v w) u (v red) u v (u v) red (u v) w, which again is what was needed to be shown If none of them are red, then all of them are blue, and we get: blue (blue blue) blue red blue red blue (blue blue) blue, which yet again is what needed to be shown Therefore, A2 holds A3: Let u,v V If u v, then clearly u v v u If u v, then one is red and the other is blue, and since red blue blue red ( blue), we have that A3 holds A4: We have already sort of seen that there is a zero vector The zero vector here is red One can check this quickly: red red red blue red blue Therefore, for all v V, v + red v Therefore red is a zero vector, and A4 holds A5: To show that A5 holds, it suffices to find, -red and -blue The element -red (if it exists) should have the property that red (-red) equals the zero element, which is also red Note that red red red Therefore choosing -red red works Similarly, -blue is an element (if any exists) such that blue (-blue) equals the zero element, red Note that blue blue red Therefore -blue blue works Therefore A5 holds M1: Let k R, v V Then k v is, by the definition, either red or blue, and is thus in V Therefore M1 holds M2: Let k R, and let u,v V Then { red if k 0, k (u v) blue otherwise k u k v { red red if k 0 blue blue otherwise red This gives us a hint which values to pick to find a counter example Pick k 1, and v u red Then, but 1 (red red) 1 red blue, 1 red 1 red blue blue red

8 Therefore, M2 fails for the specific case k 1 and u v red, and so M2 fails M3: I claim that M3 fails To show this, I need to produce two real numbers r and s, and a vector v V such that (r + s) v r v + s v Pick r 1, s 2, and v red Then, but Therefore M3 fails ( 1 + 2) red 1 red red, 1 red + 2 red blue + red blue M4: My claim is that M4 fails To see this, let r 2, s 2, and v red Then, but Therefore M4 fails 2 ( 2 red) 2 blue red, (2 2) red 4 red blue M5: This fails since for any v V, 1 v red Specifically, and therefore M5 fails 1 blue red, Therefore, A1, A2, A3, A4, A5, and M1 all hold, while M2, M3, and M4 all fail Therefore this is not a vector space

9 Question Let V be the set R 2 with the following operations defined as follows: for any (x 1,y 1 ), (x 2,y 2 ) R 2, define (x 1, y 1 ) + (x 2,y 2 ) (2(x 1 + y 1 + x 2 + y 2 ), 1(x 1 + y 1 + x 2 + y 2 )) for any k R, and for any (x 1,y 1 ) R 2, define k(x 1,y 1 ) (kx 1,ky 1 ) Is V together with these operations a vector space? If so, prove it If not, show all axioms that fail, and explain why they fail Solution To check that V is a vector space, one must check each of the 10 axioms of a vector space to see if they hold A1: Let (a,b), (c,d) V Then (a,b) + (c,d) (2(a + b + c + d), 1(a + b + c + d)) V Therefore V is closed under addition (A1 holds) A2: Let (a,b), (c,d), (e, f) V Then (a,b) + ((c,d) + (e,f)) (a, b) + (2(c + d + e + f), 1(c + d + e + f)) (2(a + b + 2(c + d + e + f) + 1(c + d + e + f)), 1(a + b + 2(c + d + e + f) + 1(c + d + e + f))) (2(a + b + c + d + e + f), 1(a + b + c + d + e + f)) ((a,b) + (c,d)) + (e,f) (2(a + b + c + d), 1(a + b + c + d)) + (e,f) (2(2(a + b + c + d) + 1(a + b + c + d) + e + f), 1(2(a + b + c + d) + 1(a + b + c + d) + e + f)) (2(a + b + c + d + e + f), 1(a + b + c + d + e + f)) Therefore this addition is associative, and so A2 holds A3: Let (a,b), (c,d) V Then (a,b) + (c,d) (2(a + b + c + d), 1(a + b + c + d)) (2(c + d + a + b), 1(c + d + a + b)) (c,d) + (a,b) Therefore A3 holds A4: I claim that A4 fails by proving the following proposition: Proposition There is no vector (x,y) V such that (1, 0) + (x,y) (1, 0) Proof If such a vector was in V, then (1,0) + (x,y) (1,0) (2(1 + x + y), 1(1 + x + y)) (1,0) (2 + 2x + 2y, 1 x y) (1,0) Therefore 2 + 2x + 2y 1 and 1 x y 0 Simplifying, we get the system 2x + 2y 1 x + y 1

10 which has no solution Therefore no such vector can exist Since the above proposition holds for (1, 0), there is no zero vector A5: If there is no zero vector, then A5 can not hope to hold M1: Let k R, (x,y) V Then Therefore M1 holds k(x,y) (kx,ky) V M2: Let k R, and let (a, b), (c,d) V Then Therefore M2 holds k((a,b) + (c,d)) k(2(a + b + c + d), 1(a + b + c + d)) (2k(a + b + c + d), k(a + b + c + d)) (2(ak + bk + ck + dk), 1(ak + bk + ck + dk)) (ak, bk) + (ck,dk) k(a,b) + k(c,d) M3: I claim that M3 fails To show this, I need to produce two real numbers r and s, and a pair (x,y) V such that (r + s)(x,y) r(x,y) + s(x,y) Let r s 1, and let (x,y) (1, 0) Then but Therefore M3 fails M4: Let r,s R, (x, y) V Then Therefore M4 holds (r + s)(x,y) (1 + 1)(1, 0) 2(1, 0) (2, 0), r(x,y) + s(x,y) 1(1, 0) + 1(1, 0) (1, 0) + (1, 0) (2( ), 1( )) (4, 2) (2, 0) r(s(x,y)) r(sx,sy) (r(sx), r(sy)) ((rs)x,(rs)y) (rs)(x,y) M5: For any (x,y) V, Therefore M5 holds 1(x,y) (1x,1y) (x,y) Thus, V is not a vector space All the axioms hold except A4, A5, and M3

= (c + a,d + b) = (c, d) + (a,b). (0, 0) + (a,b) = (0 + a,0 + b) = (a,b) (a,b) + ( a, b) = (a + a,b + b) = (0,0) = 0.

= (c + a,d + b) = (c, d) + (a,b). (0, 0) + (a,b) = (0 + a,0 + b) = (a,b) (a,b) + ( a, b) = (a + a,b + b) = (0,0) = 0. Question. Show that R 2 is a vector space. Solution. We need to check each and every axiom of a vector space to know that it is in fact a vector space. A1: Let (a,b), (c,d) R 2. Then (a,b) + (c,d) (a +